Question

How do you evaluate the integral of `int tan(x)ln(cosx) dx`?

Answer 1

`-ln^2(cosx)/2+C`

Explanation:

We can solve this integral using u-substitution, but it is fairly unapparent from the outset.

If we let `u=ln(cosx)`, differentiating this reveals that

`du=(d/dx(cosx))/cosxdx=(-sinx)/cosxdx=-tanxdx`

Multiply the interior and exterior of the integral by `-1`.

`inttanxln(cosx)dx=-int-tanxln(cosx)dx=-intudu`

Integrating this gives `-u^2/2+C`, and since `u=ln(cosx)`, this becomes `-ln^2(cosx)/2+C`.